在Python中，找到元组列表的平均值的最快方法是什么，每个元组都包含一对命名元组

Question

import numpy as numpy
from collections import namedtuple
from random import random
Smoker    = namedtuple("Smoker", ["Female","Male"])
Nonsmoker = namedtuple("Nonsmoker", ["Female","Male"])
LST = [(Smoker(random(),random()),Nonsmoker(random(),random())) for i in range(100)]

所以我有一个很长的列表，它的元素是元组。每个元组包含一对命名元组。找到这个列表的平均值最快的方法是什么？理想情况下，结果仍然是相同的结构，即(Smoker(Female=w,Male=x),Nonsmoker(Female=y,Male=z))。。

grizzly = Smoker(np.mean([a.Female for a,b in LST]),np.mean([a.Male for a,b in LST]))
panda = Nonmoker(np.mean([b.Female for a,b in LST]),np.mean([b.Male for a,b in LST]))
result = (grizzly, panda)

Answer 1

np.mean必须将列表转换为数组，这需要时间。Pythonsum节省时间：

In [6]: %%timeit
...: grizzly = Smoker(np.mean([a.Female for a,b in LST]),np.mean([a.Male for
...: a,b in LST]))
...: panda = Nonsmoker(np.mean([b.Female for a,b in LST]),np.mean([b.Male for
...:  a,b in LST]))
...: result = (grizzly, panda)
...: 
...: 
158 µs ± 597 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [9]: %%timeit
...: n=len(LST)
...: grizzly = Smoker(sum([a.Female for a,b in LST])/n,sum([a.Male for a,b in
...:  LST])/n)
...: panda = Nonsmoker(sum([b.Female for a,b in LST])/n,sum([b.Male for a,b i
...: n LST])/n)
...: result = (grizzly, panda)
...: 
...: 
46.2 µs ± 37.4 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

两者都产生相同的result(在小ε内(：

In [8]: result
Out[8]: 
(Smoker(Female=0.5383695316982974, Male=0.5493854404111675),
Nonsmoker(Female=0.4913454565011218, Male=0.47143788469638825))

如果你可以收集一个数组中的值，可能是(n，4(形状，那么平均值会很快。对于一次计算来说，这可能不值得——

In [11]: M = np.array([(a.Female, a.Male, b.Female, b.Male) for a,b in LST])
In [12]: np.mean(M, axis=0)
Out[12]: array([0.53836953, 0.54938544, 0.49134546, 0.47143788])
In [13]: timeit M = np.array([(a.Female, a.Male, b.Female, b.Male) for a,b in LST])
128 µs ± 1.22 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [14]: timeit np.mean(M, axis=0)
21.9 µs ± 371 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

由于命名元组可以像常规元组一样访问，因此我们可以直接从LST:生成数组

In [16]: np.array(LST).shape
Out[16]: (100, 2, 2)
In [17]: np.array(LST).mean(axis=0)
Out[17]: 
array([[0.53836953, 0.54938544],
[0.49134546, 0.47143788]])

但时机并不乐观：

In [18]: timeit np.array(LST).mean(axis=0)
1.26 ms ± 7.92 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

我也可以从你的列表中制作一个结构化数组-使用嵌套的dtype:

In [26]: dt = np.dtype([('Smoker', [('Female','f'),('Male','f')]),('Nonsmoker',[
...: ('Female','f'),('Male','f')])])
In [27]: M=np.array(LST,dt)
In [28]: M['Smoker']['Female'].mean()
Out[28]: 0.53836954

奇怪的是，时机相对较好：

In [29]: timeit M=np.array(LST,dt)
40.6 µs ± 243 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

但我必须分别取每个均值，否则首先将其转换为非结构化数组。

我可以用view或recfunctions实用程序从结构化数组中生成(n，4(浮点数组：

In [53]: M1 = M.view([('f0','f',(4,))])['f0']
In [54]: M1.shape
Out[54]: (100, 4)
In [55]: M2=rf.structured_to_unstructured(M)